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PHY 2053 Physics 1 Instructor: Roy Forestano Solution Worksheet 7 1 Chapter 9: Impulse and Momentum 1.1 Impulse An impulse ⃗ J supplied to an object is equivalent to an average force ⃗ F a vg acting for a time ∆ t on the object. Another definition is that the impulse ⃗ J is equal to the change in momentum of an object ∆ ⃗ p . Impulse (J): ⃗ J = area under a F vs. t graph = F avg ∆ t ⃗ J = ⃗ p f − ⃗ p i = ∆ ⃗ p Units: [ J ] = [ N ][ s ] = kg · m s 1.2 Momentum The momentum ⃗ p of an object is defined as the velocity ⃗v times the mass of the object. Momentum (p): ⃗ p = m⃗v Units: [ p ] = [ kg ] h m s i = kg · m s In a system, where no external forces are acting on an object J = 0, i.e. an isolated system, momentum is conserved. X i ⃗ P i = X f ⃗ P f 1.3 Angular Momentum We can also define a an angular impulse ⃗ J L as the average torque ⃗ τ applied to an object for a time ∆ t , or as the change in angular momentum. Angular Impulse (L): ⃗ J L = ⃗ τ ∆ t = ∆ ⃗ L Units: [ J L ] = [ L ] = kg · m 2 s

We can similarly define angular momentum ⃗ L as the product of the inertia I of the object and the angular frequency ⃗ω , Angular Momentum (L): ⃗ L = I⃗ω = m⃗v × ⃗ r = mvr Units: [ L ] = [ kg · m 2 ] 1 s = kg · m 2 s In a system, where no external torques are acting on an object J L = 0, i.e. an isolated system, angular momentum is conserved. X i ⃗ L i = X f ⃗ L f 2 Chapter 10: Energy, Work, and Power 2.1 Types of Energy Different types of linear and rotational kinetic and potential energy includes Linear Rotational K lin = 1 2 mv 2 K rot = 1 2 Iω 2 U g = mgh U rot = τθ U spring = 1 2 kx 2 U twist of spring = 1 2 κθ 2 where the units for the extra variables are [ x ] = [ m ] [ θ ] = [rad] = unitless [ v ] = h m s i [ ω ] = rad s where m is a meter, s is a second, and rad is a radian, which is unitless. The units of energy are defined as a Joule [ J ] = [ m ][ v 2 ] = h kg · m 2 s 2 i . The external work W done on an object is defined as the external force F on the object along the direction of motion times the distance d travelled, i.e. W = F · d . More generally, the net external work done W net,ext on an object is the sum of all external forces acting on the object F ext, ∥ along its direction of motion (i.e. ⃗ F ∥ ⃗ d ∥ ⃗v ) times the distance the object travels d . Work (W): W net,ext = X F ext, ∥ d From this, we can see another way to find the units of energy [ E ] = [ F ][ d ] = [ N ][ m ] = h kg · m 2 s 2 i = [ J ]. 2.2 Energy Conservation The energy conservation equation says that the total change in the internal energy ∆ E of an object is equal to the work done W on the object. ∆ E = ∆ U + ∆ K + ∆ E th + ∆ E chem = W 2

Note that the change in the internal energy ∆ E depends on the change in the kinetic energy ∆ K , change in the potential energy ∆ U , and change in other energies, such as the thermal energy ∆ E th and chemical energy ∆ E chem . Usually, when there is no mention of the thermal energy, or lost energy, and chemical energy, the energy conservation equation takes this form ∆ E = ∆ U + ∆ K = W where ∆ E th = ∆ E chem = 0 J. 2.3 Power The power P is defined as the total change in energy ∆ E = W over the total time ∆ t . An alternative definition of power P is defined as the force F ∥ times the velocity. Power (P): P = Work ∆ t = ∆ E ∆ t = F ∥ · v Units: [ P ] = [ Watt ( W )] = [ F ][ v ] = kg · m 2 s 3 3 Summary of Chapters 9 & 10 Note the three equations most used to solve the problems in these chapters can be simplified to J ext = ∆ p = ⇒ P i = P f when J ext = 0 J L,ext = ∆ L = ⇒ L i = L f when J L,ext = 0 W ext = ∆ E = ⇒ E i = E f when W ext = 0 4 Chapter 11: Applications of Energy 4.1 Laws of Thermodynamics The zeroth law of thermodynamics states that when two systems are both in thermal equilibrium with an arbitrary third system, then they are in equilibrium with each other. Roughly, T 1 = T 3 and T 2 = T 3 = ⇒ T 1 = T 2 The first law of thermodynamics states that the change in thermal energy ∆ E th of a system is equal to the work done W by the system plus the heat Q added to the system. ∆ E th = W + Q The second law of thermodynamics states that the entropy of an isolated system never decreases, i.e. ∆ S ≥ 0 The entropy must increase until the system reaches equilibrium, or stay the same if the system started in equilibrium. The entropy is at a maximum when the system is in equilib- rium. 3

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4.2 Efficiency In general, the efficiency of an engine e is defined as the work the engine puts out W out over the energy E in the engine takes in. e = what you get out what you put in = W out E in = Q H − Q C Q H where the last term comes from when you have a hot and cold reservoir; the hot reservoir expels heat out ∆ E th = Q H for the engine to perform work W and sends the rest of the heat to the cold reservoir Q C , i.e. from the first law Q H − Q C = W and the energy put in is E in = Q H . Note that the efficiency is always less than one e < 1. The maximum efficiency for an engine with a hot T H and cold reservoir T C is e max = T H − T C T H = 1 − T C T H where T is temperature in Kelvin. 4.3 Coefficient of Performance The maximum coefficient of performance COP max is defined as the inverse of the efficiency. COP max = 1 e The maximum coefficient of performance COP max,cool of a heat pump used for cooling is COP max,cool = T C T H − T C The maximum coefficient of performance COP max,heat of a heat pump used for heating is COP max,heat = T H T H − T C Note the coefficient of performance is usually larger than one, COP > 1 usually. 4

5 Exercises 5.1 Chapter 9 Exercise 1. (Knight P9.4) A raindrop with a mass of 0.014 g slams the roof at a speed of 8.1 m/s. Solution. (a) Find the impulse delivered to the roof from the raindrop. The impulse can be written as ⃗ J = ∆ ⃗ p = ⃗ F avg ∆ t . In this part, we use the fact that the impulse is the same as the change in momentum of an object. ⃗ J = ∆ ⃗ p = ⇒ J = m ( v f − v i ) = ⇒ J = (0 . 000014 kg )[(0 m/s ) − ( − 8 . 1 m/s )] = ⇒ J = 0 . 00011 kg · m/s (b) Find the magnitude of the average force of the impact if the raindrop comes to rest in 0.37 ms. We can then use the relation between the average force and the impulse to find the average force over this interval of time. ⃗ J = ⃗ F avg ∆ t = ⇒ F avg = ⃗ J ∆ t = ⇒ F avg = 0 . 00011 kg · m/s 0 . 00037 s = ⇒ F avg = 0 . 31 N Exercise 2. (Knight P9.12) A 0.43 kg soccer ball is launched at a force plate, a plate which measures the force of impact, at 16 m/s. The force versus time for the collision is shown below. Solution. 5

(a) Find the impulse. The impulse is determined by the area under the curve since this will give J = F ∆ t . J = Z Fdt = area under the curve = ⇒ J = 1 2 F max ∆ t = ⇒ J = 1 2 (0 . 007 s )(2500 N ) = ⇒ J = 8 . 75 kg · m/s (b) Find the speed the ball rebounds with. Since the impulse also determines the change in momentum and assuming the ball was initially travelling to the left, J = ∆ p = ⇒ J = m ( v f − v i ) = ⇒ v f = J m + v i = ⇒ v f = 8 . 75 kg · m/s 0 . 43 kg + ( − 16 m/s ) = ⇒ v f = 4 . 3 m/s Exercise 3. (Knight P9.24) A jellyfish with water in its stomach has a mass of 3.2 g. The jellyfish ejects 1.3 g of water and achieves a velocity of +0.070m/s. Solution. (a) Find the velocity of the ejected water, assuming the conservation of linear momentum. The initial momentum p i must be equal to the final momentum p f and since the jellyfish is at rest to start p i = 0. p i = p f = ⇒ 0 = m j v j + m w v w = ⇒ v w = m j v j m w = ⇒ v w = − (0 . 0032 kg − 0 . 0013 kg )(0 . 07 m/s ) 0 . 0013 kg = ⇒ v w = − 0 . 10 m/s Exercise 4. (Knight P9.40) A 500 g bar, with length L = 2 m and inertia I = 1 12 ML 2 , is rotating counterclockwise about its center at a rate of 120 rpm. 6

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Solution. (a) Find the angular momentum of the bar about its center. The angular momentum is defined as L = mvr = Iω , L = Iω = ⇒ L = 1 12 ML 2 ω = ⇒ L = 1 12 (0 . 5 kg )(2 m ) 2 120 rpm × 1 min 60 s × 2 πrad 1 rev = ⇒ L = 2 . 1 kg · m 2 /s (b) Now, equal amounts at each end, 0.10 m of length each with weight 25 g, of the bar are chopped off during its rotation. Using the conservation of angular momentum, find the new inertia of the rod to find its new angular speed. The new length and weight of the bar will be 1 . 8 m and 450 g, respectively. Since the angular momentum is conserved, the new angular speed will be L = I new ω new = ⇒ ω new = L I new = ⇒ ω new = L 1 12 M new L 2 new = ⇒ ω new = 2 . 1 kg · m 2 /s 1 12 (0 . 45 kg )(1 . 8 m ) 2 = ⇒ ω new = 17 . 3 rad/s Exercise 5. A 65 kg skater is twirling with their arms stretched out each at a distance of 0 . 9 m from their body with weights of 3 kg in each hand at an angular speed of 2 π rad/s. Solution. (a) Find the angular momentum of the skater assuming their body acts as a cylinder with moment of inertia I = 1 2 MR 2 where R = 0 . 25 m. Do not forget to add the moments of inertia from the weights. 7

The angular momentum is defined as L = mvr = Iω , where the moment of iner- tia of the system involves the inertia from the body and both masses. L = L f L = Iω = ⇒ L = 1 2 MR 2 + mr 2 + mr 2 ω = ⇒ L = 1 2 (65 kg )(0 . 25 m ) 2 + (3 kg )(0 . 9 m ) 2 + (3 kg )(0 . 9 m ) 2 (2 πrad/s ) = ⇒ L = 43 . 3 kg · m 2 /s (b) Assuming the skater brings the weights toward their body at a new distance of 0 . 3 m from each side, find the new angular speed of the skater. L i = L f L i = I new ω new = ⇒ ω new = L I new = ⇒ ω new = L 1 2 MR 2 + 2 mr 2 new = ⇒ ω new = 43 . 3 kg · m 2 /s 1 2 (65 kg )(0 . 25 m ) 2 + 2(3 kg )(0 . 3 m ) 2 = ⇒ ω new = 16 . 84 rad/s 8

5.2 Chapter 10 Exercise 6. (Knight P10.4) A wagon handle is being pulled at an angle of 60 o from the horizontal with a tension of 20 N. Solution. (a) Find the work done on the wagon if it is pulled 100 m at a constant speed. The only force that will contribute to the work done is the one done parallel to the displacement, i.e. the only tension component that contributes to the work is the one parallel to the surface W = Fd = T cos 60 o (100 m ) = (20 N )(100 m ) cos 60 o = 1000 J = 1 kJ Exercise 7. (Knight P10.14) An outfielder throws a 0.15 kg baseball at 32 m/s at a 30 o angle from the horizontal. Solution. (a) Find the ball’s initial kinetic energy. The initial kinetic energy only depends on the mass of the ball and its initial total speed, therefore, K i = 1 2 mv 2 i = 1 2 (0 . 15 kg )(32 m/s ) 2 = 77 J (b) Find the ball’s kinetic energy at the highest point of its motion. The kinetic energy at the balls highest point only depends on the mass of the ball and its initial horizontal velocity, v x = v i cos 30 o , as its velocity in y , or v y , at its highest point is zero, i.e v y = 0 m/s. K h = 1 2 mv 2 h = 1 2 mv 2 x = 1 2 m ( v i cos 30 o ) 2 = 1 2 (0 . 15 kg )[(32 m/s ) cos 30 o ] 2 = 58 J Exercise 8. (Knight P10.16) A wind turbine with three 45 kg blades of length 4 . 6 m each spins at 240 rpm. Solution. (a) Using the formula for rotational kinetic energy, K rot = 1 2 Iω 2 , find the kinetic energy of the blades. Assume each blade has a moment of inertia of I rod = 1 3 mL 2 . We are given ω = 240 rpm. We need to find the total inertia of the wind turbine, which will be the sum of the inertias from all three blades, i.e. I = 1 3 mL 2 + 1 3 mL 2 + 1 3 mL 2 = mL 2 . K rot = 1 2 Iω 2 = 1 2 mL 2 ω 2 = 1 2 (45 kg )(4 . 6 m ) 2 240 rev min × 1 min 60 s × 2 πrad 1 rev 2 = 300 , 000 J 9

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Exercise 9. (Knight P10.38) To be able to shift into gear and let out the clutch in order to start a manual car, the car in neutral needs an initial speed of 3.5 m/s. Solution. (a) Using the conservation of energy, find the height of an inclined hill needed to achieve exactly this speed at the bottom. Here, we assume the car begins at rest and use the work-energy theorem in the ex- change between kinetic energy and potential energy to write ∆ Ug + ∆ K = 0 = ⇒ U g,f − U g,i + K f − K i = 0. Assuming the car is initially at rest, K i = 0 and comes to the ground at its required speed, U g,f = 0, we have U g,f − U g,i + K f − K i = 0 = ⇒ − U i + K f = 0 = ⇒ U i = K f = ⇒ mgh = 1 2 mv 2 = ⇒ h = v 2 2 g = ⇒ h = (3 . 5 m/s ) 2 2(9 . 8 m/s 2 ) = ⇒ h = 0 . 63 m Exercise 10. (Knight P10.42) A 20 kg child, initially at rest, slides down a 3 . 0 m-high slide and reaches the bottom with a speed of 2 . 0 m/s. Solution. (a) Write down the initial K i and final K f kinetic energies, the initial U i and final U f potential energies, and the external work W done on the child. Note the kinetic energy is related to speed squared, and the potential energy is the gravitational potential energy here. K i = 1 2 mv 2 i = 1 2 (20 kg )(0 m/s ) 2 = 0 J K f = 1 2 mv 2 f = 1 2 (20 kg )(2 m/s ) 2 = 40 J U i = mgy i = (20 kg )(9 . 8 m/s 2 )(3 m ) = 590 J U f = mgy f = (20 kg )(9 . 8 m/s 2 )(0 m ) = 0 J W = 0 J (b) Using ∆ E = ∆ U g + ∆ K + ∆ E th = W , find the change in the thermal energy of the slide and the seat of her pants. 10

Plugging these values into the energy conservation equation, we have ∆ U g + ∆ K + ∆ E th = W = ⇒ ∆ E th = W − ∆ U g − ∆ K = ⇒ ∆ E th = 0 J − (0 J − 590 J ) − (40 J − 0 J ) = ⇒ ∆ E th = 550 J which represents the change in the thermal energy of the slide and the child’s pants. Exercise 11. (Knight P10.44) While riding a bike, person 1 pulls a trailer, with person 2 inside, up a 100 m long and 4 m high slope, where the total mass of person 2 and the trailer is 25 kg. Beginning at the bottom of the slope with a speed of 5 . 3 m/s, person 1 apples a constant force of 8 N on the trailer. Solution. (a) Using the conservation of energy equation ∆ E = ∆ U + ∆ K = W (∆ E th = 0 here), find the speed of the trailer at the top of the slope. We can find the change in energy ∆ E = W by finding the work done on the trailer W = Fd = (8 N )(100 m ) = 800 N · m and set this equal to the change in the kinetic energy plus the change in the potential energy, i.e ∆ K + ∆ U , as before, to find the final speed from the K f term. ∆ E = ∆ K + ∆ U = W = ⇒ ∆ K = W − ∆ U = ⇒ K f = W + K i − ( U f − U i ) = ⇒ 1 2 mv 2 f = Fd + 1 2 mv 2 i − ( mgh f − mgh i ) = ⇒ v f = s 2 m Fd + 1 2 mv 2 i − ( mg )( h f − h i ) = ⇒ v f = s 2 (25 kg ) (8 N )(100 m ) + 1 2 (25 kg )(5 . 3 m/s ) 2 − (25 kg )(9 . 8 m/s 2 )(4 m − 0 m ) = ⇒ v f = 3 . 7 m/s Exercise 12. (Knight P10.48) A trampoline, which can be thought of as a spring with spring constant k = 4000 N/m, initially holds a 72 kg gymnast at a depression of 0 . 80 m below equilibrium. Solution. (a) Assuming all the potential energy of the trampoline, U = 1 2 kx 2 , is transferred into their motion, find the maximum height of the gymnast. 11

We can use the conservation of energy equation in the form ∆ K + ∆ U = W = 0 J = ⇒ ∆ K + ∆ U g + ∆ U T = 0 J, where ∆ U g is the change in the gravitational potential energy and ∆ U T is the change in the trampoline’s potential energy. Note that ∆ K = 0 J since the gymnast’s velocity at the start and their highest height is 0 m/s. ∆ K + ∆ U = 0 J = ⇒ ∆ K + ∆ U g + ∆ U T = 0 J = ⇒ ∆ U g = − ∆ U T = ⇒ mgh f − mgh i = − 1 2 kx 2 f − 1 2 kx 2 i , where h i = 0 m and x f = 0 m = ⇒ h f = 1 2 kx 2 i mg = ⇒ h f = 1 2 (4000 N/m )(0 . 8 m ) 2 (72 kg )(9 . 8 m/s 2 ) = ⇒ h f = 1 . 8 m Exercise 13. (Knight 10.64) After starting from rest, a sprinter is running with a speed of 5 . 4 m/s after 1 s and 9 . 8 m/s 0 . 5 s later. Solution. (a) Find in which interval the sprinter’s output power was the largest. Remember power has units of energy ( J ) per time ( s ), or Watts, [ P ] = E t = [Watts( W )]. Here, ∆ E = ∆ K for each interval, as there is no change in the potential energy of the runner. The power in the first interval is P 1 = ∆ K 1 ∆ t 1 = 1 2 mv 2 f ∆ t 1 = 1 2 m (5 . 4 m/s ) 2 1 s = 14 . 6( m ) W/kg The power in the second interval is P 2 = ∆ K 2 ∆ t 2 = 1 2 mv 2 f − 1 2 mv 2 i ∆ t 2 = 1 2 m (9 . 8 m/s ) 2 − 1 2 m (5 . 4 m/s ) 2 0 . 5 s = 66 . 9( m ) W/kg The sprinter’s output power was largest in the second interval showing the large amount of energy required to speed the human body up. 12

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Exercise 14. (Knight 10.70) At maximum speed, a 4 kg sloth can climb a height of 6 m in 2 min. Solution. (a) Find the power output of the sloth for this climb. The energy required in the 2 min climb comes entirely from the change in the gravi- tational potential energy since the sloth starts and ends at rest, i.e. ∆ K = 0 J. Since 2 min= 120 s, we have P = ∆ U ∆ t = mg (6 m ) 120 s = (4 kg )(9 . 8 m/s 2 )(6 m ) 120 s = 1 . 96 W Exercise 15. A 20 kg cart is being pulled by a handle with a force of 30 N at an angle of 40 o from the horizontal (50 o from the vertical). Solution. (a) Find the work done by translating the cart 100 m. The only force component that contributes to the work is the one parallel to the ground W = F work d = F cos 40 o d = (30 N )(100 m ) cos 40 o = 2298 . 13 J (b) Using the work-energy theorem, find the final speed of the cart, assuming it started at rest. Using the work-energy theorem, we can equate the total work done to the total change in energy, i.e. ∆ E = ∆ U + ∆ K = W . Here, ∆ U = 0 J because the cart stays on the ground, and the initial kinetic energy K i = 0 J because the cart is initially at rest v i = 0 m/s. = ⇒ W = ∆ K + ∆ U = ⇒ W = K f − K i = ⇒ W = K f = ⇒ W = 1 2 mv 2 f = ⇒ v f = r 2 W m = ⇒ v f = s 2(2298 . 13 J ) (20 kg ) = ⇒ v f = 15 . 16 m/s 13

5.3 Chapter 11 Exercise 16. (Knight P11.6) A wind turbine extracts 40% of the kinetic energy of the wind, and 80% of this energy is converted into electricity. Solution. (a) If 110 , 000 kg of air moves past the blades at 15 m/s every second, find the power output of the generator. Hint: Use the efficiency equation e = W out E in twice to find the W gen of the generator. As the air hits the turbine, 40% of its kinetic energy is transformed into useful work, i.e. the initial work done on the turbine is e wind = W out E in = ⇒ W out = e wind E in = ⇒ W out = e wind 1 2 mv 2 = ⇒ W out = (0 . 40) 1 2 (110 , 000 kg )(15 m/s ) 2 = ⇒ W out = 4 . 95 × 10 6 J Now, this initial work from the wind is then converted into electrical energy by the generator with an efficiency of 80%. Here, E in = W out from the first part. e gen = W gen E in = ⇒ e gen = W gen W out = ⇒ W gen = e gen W out = ⇒ W gen = (0 . 80) 4 . 95 × 10 6 J = ⇒ W gen = 4 . 0 × 10 6 J The power output per second is then P = W gen ∆ t = 4 . 0 × 10 6 W. Exercise 17. (Knight 11.8) Metabolizing one molecule of glucose can form 30 molecules of ATP . Solution. (a) If the available energy in a glucose molecule is E G = 4 . 7 × 10 − 18 J and in a molecule of ATP is E AT P = 5 . 1 × 10 − 20 J, find the efficiency of this process. Use the efficiency equation where the energy in is the energy due to the metabo- lizing of a glucose molecule, i.e. E in = E G , and the creation of 30 ATP molecules is the output energy, i.e. W out = (30) E AT P . e = W out E in = (30) E AT P E G = (30)(5 . 1 × 10 − 20 J ) (4 . 7 × 10 − 18 J ) = 0 . 33 = 33% 14

Exercise 18. (Knight 11.12) Two people of mass 68 kg each are participating in two activities: running and cycling. Solution. (a) Find the metabolic energy used after 20 minutes of each activity. The metabolic power output for a 68 kg individual during a run is P r = 1150 W and that of a bike ride is P c = 480 W. Power is equivalent to energy per time, therefore, to find the metabolic energy used, we must multiply the metabolic power by the duration of 20 minutes. E r = P r ∆ t = (1150 W ) 20 min × 60 s 1 min = 1 . 4 × 10 6 J E c = P c ∆ t = (480 W ) 20 min × 60 s 1 min = 0 . 58 × 10 6 J Exercise 19. (Knight 11.14) A person of mass 75 kg raises 0 . 25 m after each pushup. Solution. (a) Write the potential energy change for one pushup ignoring the downward motion. Assume the initial height of the body is h i = 0 m. ∆ U = mg ∆ h = ⇒ ∆ U = mgh f = ⇒ ∆ U = (75 kg )(9 . 8 m/s 2 )(0 . 25 m ) = ⇒ ∆ U = 183 . 75 J This will be ∆ U = mgh f = W out . (b) Given that the human body has an efficiency of 25%, find the metabolic energy needed for one pushup. Here, ∆ U = mgh f = W out . We are looking for the metabolic energy supplied by the body to produce this motion. e = W out E in = ⇒ E in = W out e = ⇒ E in = mgh f e = ⇒ E in = 183 . 75 J 0 . 25 = ⇒ E in = 735 J 15

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(c) Find the metabolic power output for 150 pushups in 5 minutes. Note 5 min = 300 s. P = ∆ E ∆ t = (150) E in ∆ t = (150)(735 J) 300 s ≈ 370 W (d) Actually, it costs half the energy to raise the body to lower the body on each pushup. Given this, find the actual metabolic power output for 150 pushups in 5 minutes. We have to add a term to the energy that is 0 . 5 times the original energy to account for energy used in the downward motion now, i.e. P actual = ∆ E ∆ t = (150) E in + (0 . 5)(150) E in ∆ t = (1 . 5)(150) E in ∆ t = (1 . 5) (150) E in ∆ t = (1 . 5)(370 W) = 555 W Exercise 20. (Knight P11.32) A fan has a power output of 120 W. Solution. (a) Find the total energy output of the fan for 1 hour of use. E = P ∆ t = (120 W ) 1 h × 3600 s 1 h = 4 . 3 × 10 5 J Exercise 21. (Knight 11.36) A heat engine extracts 55 kJ from the hot reservoir and expels 40 kJ into the cold reservoir. Solution. (a) Find the work done W out by the engine. The work done comes from the first law of thermodynamics, E = W + Q where E = Q H and Q = Q C , E = W + Q = ⇒ W = Q H − Q C = ⇒ W = 55 kJ − 40 kJ = ⇒ W = 15 kJ (b) Find the efficiency of the heat engine. e = 1 − Q C Q H = W out Q H = 15 kJ 55 kJ = 0 . 27 16

Exercise 22. (Knight 11.38) A power plant generates 300 MW of electric power with an efficiency of 35%. Solution. (a) Using the fact that e = W out Q H and e = Q H − Q C Q H , find Q c per second, the heat exhausted to the river that cools the plant. We first need to get rid of Q H , which we do not know. Rearrange the second equation and plug in Q H = W out e from the first to get e = Q H − Q C Q H = ⇒ − Q C = eQ H − Q H = ⇒ Q C = Q H (1 − e ) = ⇒ Q C = W out e (1 − e ) = ⇒ Q C = W out 1 e − 1 = ⇒ Q C = (300 MJ/s ) 1 0 . 35 − 1 = ⇒ Q C = 560 MJ/s Exercise 23. (Knight 11.44) A power plants produces 900 MJ of electric energy per second at an efficiency of 32%. The waste heat is typically expelled to an ocean, however, it could be used to power a home requiring 20 kW for heating. Solution. (a) Find the heat expelled Q C . We can write, where Q H = Q C + W e = W Q H = W Q C + W = ⇒ e ( Q C + W ) = W = ⇒ Q C = W e − W = ⇒ Q C = W 1 e − 1 = ⇒ Q C = (900 MJ ) 1 0 . 32 − 1 = ⇒ Q C = 1 . 9 × 10 9 J 17

(b) Find the power of the expelled heat. This energy is given per second, therefore, = ⇒ P Q C = 1 . 9 × 10 9 W (c) Find how many homes can be powered by this amount of power. The power needed per home is P home = 20 kW , thus, number of homes = P Q C P home = 1 . 9 × 10 9 W 20 , 000 W = 95 , 000 homes Exercise 24. (Knight 11.50) The inside of the fridge is 0 o C and the kitchen is 20 o C. Solution. (a) Find the maximum coefficient of performance for the refrigerator. Remember for a cooling device, COP max = T C T H − T C . The temperatures in Kelvin are Q C = 0 o C +273 K = 273 K and Q H = 20 o C +273 K = 293 K. COP max = T C T H − T C = 273 K 293 K − 273 K = 13 . 6 Exercise 25. (Knight 11.56) A 60 kg person runs up 86 floors of the Empire State Building in 10 min 49 s. Solution. (a) Given that each floor is 3 . 7 m high, find the runner’s change in potential energy. ∆ U = U f − U i = (86) mgh = (86)(60 kg )(9 . 8 m/s 2 )(3 . 7 m ) = 187 kJ (b) The runner’s body has an efficiency of 25%. Find how much energy the runner’s body used during the race. The ∆ U we found in part ( a ) is the W put out by the body, thus, e = W E = ⇒ E = W e = 187 , 000 J 0 . 25 = 748 kJ (c) Find how much energy was converted to thermal energy. If the body used 25% of the energy for work, the other 75% must be the energy converted to thermal energy. E th = 748 kJ − 187 kJ = 561 kJ 18

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(d) Find the metabolic power of the runner during the run. Using the total metabolic energy during the run and dividing it by the total time of the run. Note 10 min = 600 s. P = ∆ E ∆ t = 7 . 48 × 10 5 J 649 s = 1152 . 5 W 19

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